Câu 1:
ta có \(\Delta MCD\sim\Delta MBA\) vì
\(\widehat{C}=\widehat{B}=90^o\\ \widehat{BMA}=\widehat{CMA}\)
nên \(\dfrac{MC}{MB}=\dfrac{CD}{AB}=\dfrac{6}{3}=\dfrac{8}{AB}\Rightarrow AB=\dfrac{3\cdot8}{6}=4\)
Theo đinhl lí Pi-ta-go, ta có:
\(AM^2=AB^2+BM^2\\ \Leftrightarrow AM^2=4^2+3^2\\ AM^2=25\\ AM=5\)
b) 


